Question

Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2. 65 and standard deviation. 85 (suggested in "modeling sediment and water column interactions for hydrophobic pollutants," water research, 1984: 1169–1174).

a. If a random sample of 25 specimens is selected, what is the probability that the sample average sediment density is at most 3. 00? between 2. 65 and 3. 00?

b. How large a sample size would be required to ensure that the first probability in part (a) is at least. 99?

Answer 1

I’m in your walls is your answer

Answer 2

A.

Using a standard normal distribution table, we can find that P(Z ≤ 2.06) = 0.9798.

So the probability that the sample average sediment density is at most 3.00 is approximately 0.9798 or 97.98%.

ii) To find the probability that the sample average sediment density is between 2.65 and 3.00, we can standardize the variable as follows:

z1 = (2.65 - μ) / (σ/√n) = (2.65 - 2.65) / (0.17) = 0

z2 = (3.00 - μ) / (σ/√n) = (3.00 - 2.65) / (0.17) = 2.06

Using a standard normal distribution table, we can find that P(0 ≤ Z ≤ 2.06) = P(Z ≤ 2.06) - P(Z < 0) = 0.9798 - 0.5 = 0.4798.

So the probability that the sample average sediment density is between 2.65 and 3.00 is approximately 0.4798 or 47.98%.9.

B.

P(Z ≤ z) = 0.99

Using a standard normal distribution table or a calculator, we can find that z ≈ 2.33.

Then we can solve for n as follows:

2.33 = (3.00 - 2.65) / (0.85/√n)

Simplifying this equation, we get:

n = [(2.33*0.85)/0.35]^2 ≈ 36.46

So a sample size of at least 37 is required to ensure that the probability of the sample average sediment density being at most 3.00 is at least 0.99.